Re: [Axiom-math] newbie type problem

[Martin Rubey]

Dear Frank,

 > We shall implement derivation rules via rules. First intention is:
 > ablrules := rule abl(a+b,c) == abl(a,c) + abl(b,c)
 > 
 > but axiom evaluates abl(a+b,c) to zero and makes a rule 0 ->
 > abl()+abl(). Simple ...rule 'abl(... does not work because of wrong
 > type... 

You'd have to read the section "Rules and Pattern Matching" in the axiom
book. The answer is:

abl := operator 'abl

ablplus := rule abl(a+b,c) == abl(a,c) + abl(b,c)

There is also a possibility of "adorning" pattern variables -- see the same
section. In your case, you want to have a rule that applies only to monomials,
so it will be roughly

ablmonomial := rule abl(m | monomial?(m)) == ...

However, it is pretty obvious that rules should only be the quick and dirty
approach...

Hope this helps,

Martin